Question

find D.E
x^2/a^2+y^2/b^2=1 ans. xyy''+xy'^2=yy'

Answer 1

diff. twice and eliminate a^2 and b^2 from three equations.

Answer 2

how @surjithayer

Answer 3

help please

Answer 4

Have you used the surjithayer's hint and differentiated twice ?

Answer 5

yeah

Answer 6

show you work please

Answer 7

wait

Answer 8

number 14

Answer 9

whats the derivative of \(\dfrac{x^2}{a^2}\) with respect to \(x\) ?

Answer 10

but i think , may solution is not right

Answer 11

yes

Answer 12

Keep in mind, \(a\) and \(b\) are constants here

Answer 13

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\]

differentiating implicitly with respect to \(x\) gives
\[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0\tag{2}\]

yes ?

Answer 14

okay :)

Answer 15

but , i did it first , but my answer was not right

Answer 16

your work in that picture is wrong
you're not treating a^2 as constant, instead trying to differentiate it and writing 2a.. which is wrong

Answer 17

can you help please

Answer 18

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\]

differentiating implicitly with respect to \(x\) gives
\[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'} \]

Plug this in first equation and get
\[\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1 \]
\[\implies -xyy' +y^2 = b^2 \tag{2}\]

differentiate this again and you're done!

Answer 19

@ganeshie8 gets :)) thank you very much :))

Answer 20

yw, are you able to differentiate it again and arrive at the answer ?

Answer 21

@ganeshie8 im very confused

Answer 22

remember product rule ?

Answer 23

yeah i remember

Answer 24

how about the b^2 ?

Answer 25

b^2 is just a constant, like, 3^2 or 5^2

Answer 26

it doesn't change, so the derivative is 0

Answer 27

\[-x y y'' . -1 + x y^2.y'=b^2\]

Answer 28

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\]

differentiating implicitly with respect to \(x\) gives
\[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'} \]

Plug this in first equation and get
\[\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1 \]
\[\implies -xyy' +y^2 = b^2 \tag{2}\]

differentiating w.r.t \(x\) again
\[-(yy' +y'xy' + y''xy) + 2yy'= 0\]
\[yy' = xyy'' + x{y'}^2\]

Answer 29

Recall below product rules
\[\left(f(x)*g(x)\right)' = f'(x)g(x) + f(x)g'(x)\]

\[\left(f(x)*g(x)*h(x)\right)' = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)\]

Answer 30

@ganeshie8 i got the answer :))))

Answer 31

how about
x^2+axy+bx+c=0

Answer 32

Give it a try first

Answer 33

i have a answer

Answer 34

@ganeshie8 number 11

Answer 35

Looks you forgot to use product rule here ?

Answer 36

isn't that we have to find the second and third derivatives and then plug them back into the equation to check if the equation we are given (that ans. thing) holds true?

Answer 37

@ganeshie8

Answer 38

x and y's are variables
treat a b c as constants which when the derivative is taken it goes to 0

Answer 39

ai derivative of x is 1 = a (1) ?

Answer 40

@UsukiDoll this is like constructing the differential equation from the given general solution by eliminating the arbitrary constants

Answer 41

oh I see so we're not verifying... we're finding a DE

Answer 42

yeah we can think of the equations of given ellipses/parabolas as general solutions

Answer 43

yeah , were finding D.E

Answer 44

@jacalneaila did you find ur mistake in #11 ?

Answer 45

yeah, i try to solve ut again , but may answer is different now :(

Answer 46

*it

Answer 47

show ur work

Answer 48

wait , i dont have a cellphone

Answer 49

#11 should be fairly simple compared to the earlier one

Answer 50

\[ax+yy'+a(1)? \]

Answer 51

@ganeshie8